What is (\partial)/(\partial x)(2(xe^{x/2+y}+1)) ?

The answer to (\partial)/(\partial x)(2(xe^{x/2+y}+1)) is e^{(x+2y)/2}x+2e^{x/2+y}

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(\partial)/(\partial x)(2(xe^{x/2+y}+1))

\frac{\partial}{\partial x} (2 (x e^{\frac{x}{2} + y} + 1))

e^{\frac{x + 2 y}{2}} x + 2 e^{\frac{x}{2} + y}

Solution steps

\frac{\partial}{\partial x} (2 (x e^{\frac{x}{2} + y} + 1))

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{\partial}{\partial x} (x e^{\frac{x}{2} + y} + 1)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= 2 (\frac{\partial}{\partial x} (x e^{\frac{x}{2} + y}) + \frac{\partial}{\partial x} (1))

\frac{\partial}{\partial x} (x e^{\frac{x}{2} + y}) = e^{\frac{x}{2} + y} + \frac{e ^{\frac{x + 2 y}{2}} x}{2}

\frac{\partial}{\partial x} (1) = 0

= 2 (e^{\frac{x}{2} + y} + \frac{e ^{\frac{x + 2 y}{2}} x}{2} + 0)

Simplify

2 (e^{\frac{x}{2} + y} + \frac{e ^{\frac{x + 2 y}{2}} x}{2} + 0) : e^{\frac{x + 2 y}{2}} x + 2 e^{\frac{x}{2} + y}

= e^{\frac{x + 2 y}{2}} x + 2 e^{\frac{x}{2} + y}

\frac{d}{d x} (2 e^{x} + 3 x - ln (x))

\int - \frac{1}{2} sin (2 x) d x

x \to 80 lim (82 - x)

\int 2 sin (π x) d x

x^{4} - 6 x^{3}

What is (\partial)/(\partial x)(2(xe^{x/2+y}+1)) ?
The answer to (\partial)/(\partial x)(2(xe^{x/2+y}+1)) is e^{(x+2y)/2}x+2e^{x/2+y}